similar to: lm without intercept

The data set consists of two sets of matrices, as labelled by the columns, T's and C's. > xy x T1 T2 T3 T4 T5 C1 C2 C3 C4 C5 [1,] 50 0.00 0.00 33.75 0.00 0.00 0.00 36.76 0.00 35.26 0.00 [2,] 13 34.41 0.00 0.00 36.64 32.86 34.11 35.80 37.74 0.00 0.00 [3,] 14 35.85 0.00 33.88 36.68 34.88 34.58 0.00 32.75 37.45 0.00 [4,] 33 34.56

= numbersToWords - convert your integers and floats to english or japaneese words,or translate it into american currency - This plugin was created from a solution to the ruby quiz (english numbers) by Matthew D Moss. I modified the code to work as a rails plugin and added the to_dollars method. - the to_dollars method will work with integers and floats - the to_english and to_japaneese methods

Hi, I tried to inverse a matrix but it doesn`t work. I hope somebody can help me. This is what I did. > kurse <- read.table("kurse.txt", header=T, dec=",") > x <- cbind(1,kurse[,-c(1)]) > y <- kurse$index > t(x) %*% x Error in t(x) %*% x : requires numeric matrix/vector arguments > x <- as.matrix(x) > xtxi <- solve(t(x) %*% x) Error in

I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24 14.7 ... $ I002N2: num 2.92 2.54 0.11 1.6 7.12 ... $ I003N2: num -22.362 -0.222 -19.291

While trying to train randomForest with my dataset, I am ending up with the following error Error in randomForest.default(datatrain, classtrain) : length of response must be the same as predictors My data looks like: A,B,C,D,Class 1,2,1,2,cl1 1,2,1,2,cl1 3,2,1,2,cl2 3,2,1,2,cl2 3,2,1,2,cl2 3,2,1,2,cl2 3,2,1,2,cl2 3,2,1,2,cl2 3,2,1,2,cl2 3,2,12,3,cl2 3,2,1,2,cl2 Actual dataset has around 4000

Hello, This is a standard example in which I read the time series data from a csv file and create a zoo object: x0 <- read.csv(file="CPI.csv", header=TRUE) time_0<-as.yearmon("1981-01")+(0:371)/12 x0zoo<-zoo(x0, time_0) The data look like this: TIME CPI CPI_food CPI_Clothes CPI_House CPI_Rent 198101 62.1 55.34 103.45 65.24 61.43 198102 63.16 56.95

Dear Rxperts, q <- data.frame(p=rep(c("A","B"),each=10,len=30), a=rep(c(1,2,3),each=10),id=seq(30), b=round(runif(30,10,20)), c=round(runif(30,40,70))) The operation below... tabular(((p=factor(p))*(a=factor(a))+1) ~ (N = 1) + (b + c)* (mean+sd),data=q) yields some rows of NAs and NaN as shown below b c p a N mean sd mean sd A 1 10

Hi, I want to test for a contrast from a regression where I am regressing the columns of a matrix. In short, the following. X <- matrix(rnorm(50),10,5) Y <- matrix(rnorm(50),10,5) lm(Y~X) Call: lm(formula = Y ~ X) Coefficients: [,1] [,2] [,3] [,4] [,5] (Intercept) 0.3350 -0.1989 -0.1932 0.7528 0.0727 X1 0.2007 -0.8505 0.0520

Dear All, I try to use Sweave to make a document. But when I use the Sweave() command on it and build a pdf with pdflatex (3.141592-1.40.3) my apostrophes are replaced by some gibberish (an 'a' with a hat on it, a capital A with a arc pointing upwards on it and a capital Y with two points on it). If I manually replace the apostrophes using the keyboard, I get a different looking

Hello, I was doing a linear regression with the following formula: *lm(y~x+0)*, so it passes through the origin. But when I called the summary of the regression i saw that R squared is abnormally high (it's a lot lower in other programs such as SigmaPlot and MS Excel).The manual explained the cause of the difference (because of the different computing method), but what should I do to get the

Hello, I was doing a linear regression with the following formula: lm(y~x+0), so it passes through the origin. But when I called the summary of the regression i saw that R squared is abnormally high (it's a lot lower in other programs such as SigmaPlot and MS Excel).The manual explained the cause of the difference (because of the different computing method), but what should I do to get the

Hi, I've tired to forward a Fax from Asterisk to Hylafax. It works so far until I tried with a Fax machine. I just got error shown in the log below. I'm not sure why. I've tested it with other 6 machines and they all work fine. Do you have any idea why? Pim Hylafax Session log: Apr 12 11:16:48.82: [ 5933]: SESSION BEGIN 000000078 492212601860 Apr 12 11:16:48.82: [ 5933]:

Does something like this get you close: x <- list() keys <- LETTERS[1:6] # create for (i in keys){ x[[i]] <- data.frame(a=1:5, b=1:5, c=1:5) } # output output <- file('tempxx.txt', 'w') for (i in keys){ write.table(i, row.names=FALSE, col.names=FALSE, file=output, quote=FALSE) write.table(x[[i]], file=output, quote=FALSE) } close(output) On Wed, Jul 2,

Hello, Using the following lines of code, I created the following graph:

Hello, Using the following lines of code, I created the following graph:

hello How can I change a vector's float value from ("1.085714e-01") format to a simpler format (like 0.1xxx). regards -- Ahmet Temiz Jeo. M?h. Afet ??leri Gen. Md.l??? Deprem Ar. D. Ahmet Temiz Geo. Eng. General Dir. of Disaster Affairs -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean.

I have used the greedy.wilks to stepwise discriminant analysis, but it doesn't work with my dataset. I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24

Hello, I need help with a partial least square regression in R. I have read both the vignette and the post on R bloggers but it is hard to figure out how to do it. Here is the script I wrote: library(pls) plsrcue<- plsr(cue~fb+cn+n+ph+fung+bact+resp, data = cue, ncomp=7, na.action = NULL, method = "kernelpls", scale=FALSE, validation = "LOO", model = TRUE, x = FALSE, y =

Dear Users, In case of ridge logistic regression, i want to calculate the optimum penalty using aic and bic criteria. Here is the sample code: fit <- lrm(RES ~CAT01+NUM01+NUM02+CAT02+CAT03+CAT04+NUM03+CAT05+CAT06+NUM04+ CAT07+CAT08+NUM05+NUM06, data = train.data, x = TRUE, y = TRUE) pentrace(fit, penalty = list(seq(.001, 5, by=.1))) output: Best penalty: penalty df 1.001

Version 3.9 of the survey package is now on CRAN. Since the last announcement (version 3.6-11, about a year ago) the main changes are - Database-backed survey objects: the data can live in a SQLite (or other DBI-compatible) database and be loaded as needed. - Ordinal logistic regression - Support for the 'mitools' package and multiply-imputed data - Conditioning plots,